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November 15th, 2007, 02:12 AM  #1 
Newbie Joined: Nov 2007 Posts: 3 Thanks: 0  Constructing a formula
I always knew how to do this, until yesterday, my mind just went blank. I'm writing a computer program which will ask questions on constructing a formula... I need to know how to write down the working to find this out. This has been bothering me. Thanks to all who contributes, Garry. 
November 15th, 2007, 02:51 AM  #2 
Senior Member Joined: May 2007 Posts: 402 Thanks: 0 
What kind of formula and what kind of questions?

November 15th, 2007, 03:03 AM  #3 
Newbie Joined: Nov 2007 Posts: 3 Thanks: 0 
Oh yeah, oops. Forgot to say what kind of formula. How do you find out that formula. M  1 2 3 4 5 6 7 8 Y  6 9 12 15 18 Y = 2M + 5 
November 15th, 2007, 03:29 AM  #4 
Senior Member Joined: May 2007 Posts: 402 Thanks: 0 
Ok, so what you need is a line. First, let me say that you can make a line go through exactly 2 random points, or any number of specific points. So, the information you gave is too much  you only need two points. So, let's say you have these points: T1(x1,y1) T2(x2,y2) and you need an equation of the form: y(x)=a*x+b So, plugging in the two points you get: y(x1)=y1=a*x1+b y(x2)=y2=a*x2+b  subtract the second one from the first y1y2=a*x1+ba*x2b y1y2=a(x1x2)  solve for 'a' a=(y1y2)/(x1x2)  plug a into the first (or the second) formula y1=(y1y2)/(x1x2)*x1+b  solve for b b=y1(y1y2)/(x1x2)*x1 b=y1(y1*x1y2*x1)/(x1x2) b=(y1*x1y1*x2y1*x1+y2*x1)/(x1x2) b=(y2*x1y1*x2)/(x1x2)  plug that and 'a' into y(x)=a*x+b y(x)=(y1y2)/(x1x2)x+(y2*x1y1*x2)/(x1x2) y(x)=(y1y2)/(x1x2)x+(y2*x1+y2*x2y2*x2y1*x2)/(x1x2) y(x)=(y1y2)/(x1x2)x+(y2*x1y2*x2)/(x1x2)+(y2*x2y1*x2)/(x1x2) y(x)=(y1y2)/(x1x2)x+y2+(y2y1)/(x1x2)x2 y(x)y2=(y1y2)/(x1x2)(xx2)  or if we had plugged it into the second equation y(x)y1=(y1y2)/(x1x2)(xx1) but both of them yield the same result! 
November 15th, 2007, 03:43 AM  #5 
Newbie Joined: Nov 2007 Posts: 3 Thanks: 0 
Made no sense to me. Not what my teacher taught me. It was way simpler. I'm sorry I didn't understand. That must of took a while, but thank you for your help anyway. 
November 15th, 2007, 05:41 AM  #6  
Senior Member Joined: Dec 2006 Posts: 1,111 Thanks: 0 
I'm not sure I would have done it quite like that. Since we are trying to find a linear line, all we need are two points, like you said, but I would find the equation a little differently: (Maybe "milin" could explain further why he used the method he did, as he must have had a reason). Quote:
First, we find the slope between the first two points, (1,6) and (2,9): Slope = (96)/(21) = 3 y = 3M + b (new equation) Now, plug in any point (We'll use 1,6) and find the value of 'b': 6 = 3(1) + b 3 = b y = 3M + 3 Final equation of the line which goes through all the points you originally mentioned. (I'm not sure where you got the "y = 2M + 5" equation from, but it does not work)  

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